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Date	November 2018	Marks available	3	Reference code	18N.1.AHL.TZ0.H_11
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Find	Question number	H_11	Adapted from	N/A

Question

Let S be the sum of the roots found in part (a).

Find the roots of ${z^{24}} = 1$ which satisfy the condition $0 < {\text{arg}}\left( z \right) < \frac{\pi }{2}$ , expressing your answers in the form $r{e^{{\text{i}}\theta }}$ , where $r$ , $\theta \in {\mathbb{R}^ + }$ .

[5]

Show that Re S = Im S.

[4]

b.i.

By writing $\frac{\pi }{{12}}$ as $\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right)$ , find the value of cos $\frac{\pi }{{12}}$ in the form $\frac{{\sqrt a + \sqrt b }}{c}$ , where $a$ , $b$ and $c$ are integers to be determined.

[3]

b.ii.

Hence, or otherwise, show that S = $\frac{1}{2}\left( {1 + \sqrt 2 } \right)\left( {1 + \sqrt 3 } \right)\left( {1 + {\text{i}}} \right)$ .

[4]

b.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\left( {r\left( {{\text{cos}}\,\theta + {\text{i}}\,{\text{sin}}\,\theta } \right)} \right)^{24}} = 1\left( {{\text{cos}}\,0 + {\text{i}}\,{\text{sin}}\,0} \right)$

use of De Moivre’s theorem (M1)

${r^{24}} = 1 \Rightarrow r = 1$ (A1)

$24\theta = 2\pi n \Rightarrow \theta = \frac{{\pi n}}{{12}}$ , $\left( {n \in \mathbb{Z}} \right)$ (A1)

$0 < {\text{arg}}\left( z \right) < \frac{\pi }{2} \Rightarrow n =$ 1, 2, 3, 4, 5

$z = {\text{e}}\frac{{\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{2\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{3\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{4\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{5\pi {\text{i}}}}{{12}}$ A2

Note: Award A1 if additional roots are given or if three correct roots are given with no incorrect (or additional) roots.

[5 marks]

Re S = ${\text{cos}}\frac{\pi }{{12}} + {\text{cos}}\frac{{2\pi }}{{12}} + {\text{cos}}\frac{{3\pi }}{{12}} + {\text{cos}}\frac{{4\pi }}{{12}} + {\text{cos}}\frac{{5\pi }}{{12}}$

Im S = ${\text{sin}}\frac{\pi }{{12}} + {\text{sin}}\frac{{2\pi }}{{12}} + {\text{sin}}\frac{{3\pi }}{{12}} + {\text{sin}}\frac{{4\pi }}{{12}} + {\text{sin}}\frac{{5\pi }}{{12}}$ A1

Note: Award A1 for both parts correct.

but ${\text{sin}}\frac{{5\pi }}{{12}} = {\text{cos}}\frac{\pi }{{12}}$ , ${\text{sin}}\frac{{4\pi }}{{12}} = {\text{cos}}\frac{{2\pi }}{{12}}$ , ${\text{sin}}\frac{{3\pi }}{{12}} = {\text{cos}}\frac{{3\pi }}{{12}}$ , ${\text{sin}}\frac{{2\pi }}{{12}} = {\text{cos}}\frac{{4\pi }}{{12}}$ and ${\text{sin}}\frac{\pi }{{12}} = {\text{cos}}\frac{{5\pi }}{{12}}$ M1A1

⇒ Re S = Im S AG

Note: Accept a geometrical method.

[4 marks]

b.i.

${\text{cos}}\frac{\pi }{{12}} = {\text{cos}}\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) = {\text{cos}}\frac{\pi }{4}{\text{cos}}\frac{\pi }{6} + {\text{sin}}\frac{\pi }{4}{\text{sin}}\frac{\pi }{6}$ M1A1

$= \frac{{\sqrt 2 }}{2}\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 2 }}{2}\frac{1}{2}$

$= \frac{{\sqrt 6 + \sqrt 2 }}{4}$ A1

[3 marks]

b.ii.

${\text{cos}}\frac{{5\pi }}{{12}} = {\text{cos}}\left( {\frac{\pi }{6} + \frac{\pi }{4}} \right) = {\text{cos}}\frac{\pi }{6}{\text{cos}}\frac{\pi }{4} - {\text{sin}}\frac{\pi }{6}{\text{sin}}\frac{\pi }{4}$ (M1)

Note: Allow alternative methods eg ${\text{cos}}\frac{{5\pi }}{{12}} = {\text{sin}}\frac{\pi }{{12}} = {\text{sin}}\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right)$ .

$= \frac{{\sqrt 3 }}{2}\frac{{\sqrt 2 }}{2} - \frac{1}{2}\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 - \sqrt 2 }}{4}$ (A1)

Re S = ${\text{cos}}\frac{\pi }{{12}} + {\text{cos}}\frac{{2\pi }}{{12}} + {\text{cos}}\frac{{3\pi }}{{12}} + {\text{cos}}\frac{{4\pi }}{{12}} + {\text{cos}}\frac{{5\pi }}{{12}}$